Whakaoti i te 2xx-5x=0

$$2 x x - 5 x = 0$$

Whakaoti mō x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te x.

$$x\left(2x-5\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2x-5=0.

$$x=0 x=\frac{5}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia te x ki te x, ka x^{2}.

$$2x^{2}-5x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -5 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2\times 2}$$

Tuhia te pūtakerua o te \left(-5\right)^{2}.

$$x=\frac{-\left(-5\right)±5}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±5}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{5±5}{4}$$

Nā, me whakaoti te whārite x=\frac{5±5}{4} ina he tāpiri te ±. Tāpiri 5 ki te 5.

$$x=\frac{10}{4}$$

Whakahekea te hautanga \frac{10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±5}{4} ina he tango te ±. Tango 5 mai i 5.

$$x=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{5}{2} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia te x ki te x, ka x^{2}.

$$2x^{2}-5x=0$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-5x}{2}=\frac{0}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}-\frac{5}{2}x=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$x^{2}-\frac{5}{2}x=0$$

Whakawehea te -\frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{4}. Nā, tāpiria te pūrua o te -\frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Pūruatia -\frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{25}{16}$$

Tauwehea te x^{2}-\frac{5}{2}x+\frac{25}{16}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$x-\frac{5}{4}=\frac{5}{4} x-\frac{5}{4}=-\frac{5}{4}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

$$x=\frac{5}{2} x=0$$